Physics 101 t11

Physics 101 t11 - FORM 0 - PR EME mls mls [D] 0.58 x 104...

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Unformatted text preview: FORM 0 - PR EME mls mls [D] 0.58 x 104 mls 1.C W is c o r r e c t l y e x p r e s s e d as [B] 5.80 x 10 3 [C] 5.8 x 10 3 OM The quotient, 5800 mls x 10 3 D41 1.C OM P h y s i c s 101 D41 [2] T w o a t h l e t e s j u m p s t r a i g h t up. J o h n h a s twice t h e i n i t i a l s p e e d of Harry. C o m p a r e d t o @ w i c e as l o n g 1.C OM - PR EME Harry, J o h n s t a y s i n t h e a i r [A] 0.50 t i m e s as long [ B ] 1 . 4 1 t i m e s as long [D] t h r e e t i m e s as long [E] f o u r t i m e s as long 1.C OM - PR EME D41 [3] I f you a r e walking 3 m i n o r t h , t h e n 6 m i west, a n d t h e n 5 m i n o r t h , w h a t is t h e m a g n i t u d e of your t o t a l displacement? [A] 3 m i [B) 6 m i [CJ 8 m i (@J10 mi [E]14 mi 11. COM - PR EME D41 [4J A p e r s o n t h r o w s a b a l l h o r i z o n t a l l y from t h e t o p o f a b u i l d i n g t h a t is 20.0 m h i g h f r o m t h e g r o u n d level. T h e b a l l l a n d s 50.0 m d o w n r a n g e f r o m t h e b a s e o f t h e b u i l d i n g . W h a t was t h e i n i t i a l v e l o c i t y o f t h e b a l l ? 0 2 4 . 8 mls [B] 49.5 mls [C] 99.0 m l s [D] 148 m l s [E] 202 mls COM - PR EME D4 [5] A 2 0 - t o n t r u c k collides w i t h a 1500-lb c a r a n d causes a l o t o f d a m a g e t o t h e car. Since a l o t o f d a m a g e is done o n t h e c a r [A] t h e force o n t h e t r u c k is g r e a t e r t h e n t h e force on t h e c a r ([:i3D t h e force o n t h e t r u c k is equal t o t h e force o n t h e c a r t h e force o n t h e t r u c k is s m a l l e r t h a n t h e force on t h e c a r [D] t h e t r u c k d i d n o t slow down d u r i n g t h e collision [E] t h e c a r did n o t slow down d u r i n g t h e collision COM - PR EME D4 11. 1Cl COM - PR EM ED4 11. [6] A 25-kg c r a t e is b e i n g p u l l e d along a h o r i z o n t a l s m o o t h surface. T h e p u l l i n g force is 20.0 N a n d is d i r e c t e d 20.0° above t h e h o r i z o n t a l . W h a t is t h e m a g n i t u d e of t h e a c c e l e r a t i o n o f the crate? [A] 0.26 m/s 2 [Bj 0.30 m/s 2 [C] 0.44 m/s 2 [D] 0.64 m / s 2 @ 0 . 7 5 m/s 2 11. COM - PR EM ED4 11. [7] A 1250-kg c a r is picking u p s p e e d as i t goes a r o u n d a h o r i z o n t a l curve whose r a d i u s is 100.0 m . T h e coefficient o f s t a t i c f r i c t i o n b e t w e e n t h e t i r e s a n d t h e r o a d is 0.35. At w h a t will t h e c a r b e g i n t o skid sideways? mls [B] 23.6 mls [C] 29.4 mls 34.3 mls [E] 35.0 mls ED4 11. COM - PR EM ED4 [8] A LO-kg o b j e c t moving i n a c e r t a i n d i r e c t i o n h a s a k i n e t i c energy o f 2.0 J . I t h i t s a wall a n d comes b a c k w i t h h a l f i t s o r i g i n a l s p e e d . W h a t is t h e k i n e t i c e n e r g y o f t h i s o b j e c t a t this point? [A] 2.0 J [B]LO J @0.50 J [D] 0.25 J [El 4.0 J PRE MED 411 .CO MPRE M ED4 11. COM - PR EM ED4 11. COM - PR EM [9] A 40.0-N h o r i z o n t a l force p u s h e s a n o b j e c t along a r o u g h floor so t h a t t h e o b j e c t moves w i t h a c o n s t a n t velocity of 4.00 m / s . How m u c h p o w e r does t h i s force deliver t o t h e o b j e c t ? [A] 0 W W [CJ 320 W [DJ 480 W [El 960 W P h y s i c s 101 OM FORM 0 - PR EME D41 1.C OM - PR EME D41 1.C [10] A force of 30.0 N a p p l i e d t o a s p r i n g s t r e t c h e s i t b y 4.0 cm. W h a t is t h e p o t e n t i a l e n e r g y of t h e s p r i n g i n t h e s t r e t c h e d p o s i t i o n ? [Al 0.20 J [Bl 0.30 J [e] DAD J @>o.60 J [El 1.2 J i f..0: : - 0 ) 1 ) U.t::' - PR EME v:, -= ViD 3, l D41 1.C OM - PR EME D41 1.C OM [11] A bicyclist is finishing his r e p a i r o f a flat t i r e w h e n a f r i e n d r i d e s by a t 3.2 m/s. T h r e e (3.0) seconds l a t e r , t h e b i c y c l i s t h o p s o n his b i ke a n d a c c e l e r a t e s a t 204 m/s 2 u n t i l h e c a t c h e s his m e n d . ( a ) How m u c h t i m e d o e s i t t a k e u n t i l h e c a t c h e s his f r i e n d ? ( b ) How f a r h a s he t r a v e l e d in t h i s t i m e ? ( c ) W h a t is h i s s p e e d w h e n h e c a t c h e s u p ? J V.,lOS, XJ.D-=D X4 ==X,o · t if, t:= lJ; ( .t t 3.Dlo) J 11. COM x/()-=. - PR EME D4 + ) , . '1 t lA.. 3; l t -+ - PR EME D4 ,::: 11. ED4 /3;1..1...-+ '1·-1.1 2. t, 2. 1- t/<..) 4/ - = = ) f,:( lj"t - PR EM t -. . . 3,1... i COM ':=. ::XlO + S _ 11. - PR EM - PR EM COM 11. ED4 - PR EM COM 11. ED4 - PR EM COM 11. ED4 MPRE M .CO 411 3. ( 4l T 3, 0) M ::: 2.. i f ED4 -= (. Lr1.. V + C( t MED l.I) - 11. COM v-; ( tt 3. 0-:;.) =- til. PRE i . - 3, l t -- 5,t = D $ t-, S st ED4 t. Xl -:... 1-. 4 XL': Cc) ttctz f/< ' COM + .' q. i t t s (4. Ss.) . 1. - ' (b) I ( 11. I Ilj Xl COM t= ? Xt = Xt (0) -2,1; ist! 1.. 4 . C,I. 4 ' . :::s :;: I O.f. /s (r ( . I '1 FORM 0 OM P h y s i c s 101 - PR EME D41 1.C OM - PR EME D41 1.C OM - PR EME D41 1.C OM - PR EME D41 1.C [12] T w o blocks a r e c o n n e c t e d b y a massless s t r i n g , a s s h o w n i n t h e d r a w i n g ( t h e p u l l e y is a s s u m e d t o b e massless a n d frictionless). T h e frictionless i n c l i n e d s u r f a c e makes a n a n g l e () = 36° w i t h t h e h o r i z o n t a l . Block 1 o n t h e i n c l i n e h a s a m a s s m l = 5.5 kg, a n d t h e h a n g i n g b l o c k 2 h a s a m a s s m 2 = 4.5 kg. ( a ) F i n d t h e m a g n i t u d e a n d d i r e c t i o n o f t h e a c c e l e r a t i o n o f block 2. ( b ) F i n d t h e t e n s i o n i n t h e s t r i n g c o n n e c t i n g t h e two blocks. (c) W h a t s h o u l d t h e a n g l e () b e c h a n g e d t o ( w h i l e l e a v i n g m l a n d m 2 u n c h a n g e d ) i n o r d e r for the system of masses to be in equilibrium? )11,4 I (>tI,t - C{:; - PR EME D4 11. COM -= ( 1\-1<.- I kt (j COM 1-rI1 + VII t 11. .. ; - i1 /:.Sot COM , PRE MED 411 .CO MPRE M ED4 11. COM - PR EM ED4 11. COM - PR EM ED4 11. COM - PR EM ED4 11. COM - PR EM ED4 11. COM - PR EM ED4 11. r,\ . ::: l , Z V - PR EME D4 S I (t. ,fYsL) • P h y s i c s 101 1.C OM FORM 0 -= 12 S6k. () I 6 :::. h::: I t. () kA J - PR EME D41 G 1.C f:- -= 6 t<. + . 6 U. J Wf- t W-e I 6.U = \!Jj/\t: -= OM tel) - PR EME \4\ D41 1.C OM - PR EME D41 1.C OM - PR EME D41 [13] A 1250-kg c a r drives u p a hill t h a t is 17.0 m h i g h . D u r i n g t h e d r i v e , two n o n c o n s e r v a t i v e forces d o w o r k o n t h e c a r : ( i ) t h e f o r c e o f k i n e t i c f r i c t i o n , a n d ( i i ) t h e f o r c e g e n e r a t e d b y t h e c a r ' s e n g i n e . T h e work d o n e by k i n e t i c f r i c t i o n is - 3 . 5 0 X 10 5 J; t h e work d o n e by t h e e n g i n e is + 4 . 7 0 x 1 0 5 J . F i n d ( a ) t h e c h a n g e i n t h e c a r ' s k i n e t i c e n e r g y f r o m t h e b o t t o m t o t h e t o p o f t h e hill, ( b ) t h e s p e e d o f t h e c a r w h e n i t r e a c h e s t h e t o p o f t h e h i l l , i f i t s s p e e d a t t h e b o t t o m is 18.0 mIs, (c) t h e average m a g n i t u d e o f t h e k i n e t i c f r i c t i o n force a c t i n g o n t h e c a r , i f t h e l e n g t h o f t h e j o u r n e y u p t h e h i l l i s 85.0 m . I COM =: 11. 0f J - PR EME D4 (b) lrL, := (g,o COM - PR EME D4 11. b k =- W\1 '-- 6 L{::- Wf Wt - h-r J6. . b k =- ( LI.7-0- 3.so) ' / 0 ) J - 12'5'0 ':Lllt·/'.O J =- - 8, 2..)' _ 'l."" _{f6k . '\,..:l 6 k =. 1"'" (Vi L_tz· - ) V{ - ) ' V \ + - PR EM ED4 11. f a· PRE MED 411 .CO MPRE M ED4 11. COM - PR EM ED4 11. COM - PR EM ED4 11. COM - PR EM ED4 11. COM - PR EM ED4 11. COM 1F(- ::. COM 1 f l g.OL .= 1'). s V( J . , IO't:s· FORM 0 OM P h y s i c s 101 - PR EME D41 1.C FORMULAE SHEET A cos 0A, Ay = i B 1.C a - PR EME Newton's Laws: mg( -y), - PR EME D4 = FdcosB, K W = Flld COM Ug = m g y , MPRE M ED4 11. COM - PR EM ED4 11. COM - PR EM ED4 11. COM - PR EM ED4 11. COM - PR EM ED4 11. COM - PR EM ED4 11. COM - PR EME D4 11. We = - D . U , .CO kinetic: i k Vo (t - 0) , - .. V.. v; = Vx - v o ; + v . , 2 + 2a",(:c R= J.LkN, :co) s i n ( 2Bo ) 9 s t a t i c : i s ::; J.LsN . m i. cp = - ,v.2 ( - TA) ,. ( - f ) , circular motion: Work and Energy: 411 4:- D.r= rf - ri g( -yL 11. COM friction: S p r i n g s : Fx = - k : c , MED tanBA = D41 1.C OM Free fall and projectile motion: PRE i + (-B), - PR EME D41 Motion at constant acceleration: :Co :c(t 0), Vx = vOx + a x t , :c = :Co + i ! x t , :c = :Co + v o " , t + W= JA'; + OM Relative velocity: Weight: 2a AsinOA, A = - PR EME i + B = (Ax + B x ) x + (Ay + By)Y, 1.C = A x x + AyY, A x D41 i Vectors: :c= OM Quadratic equation: = D.K, +U, W P=t = Flli! ...
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This note was uploaded on 10/17/2011 for the course PHY 101 taught by Professor Ashkenkai during the Fall '08 term at FIU.

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